I Definition of a group

  • Thread starter Lebnm
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In general, the textbooks says that, if the set ##G## is a group, so to every element ##g \in G## there is other element ##g^{-1} \in G## such that ##g g^{-1} = g^{-1}g = e##, where ##e## is the identity of the group. But I am reading a book where this propriete is write only as ##g^{-1} g = e##, and the book says that ##g g^{-1} = e## follows from this. The proof it gives is: Applaying ##g## on the left on the both sides, we have $$g(g^{-1} g) = (g g^{-1}) g = g e = g,$$and of this the book concludes that ##g g^{-1}## is equal to ##e##, because its action in ##g## gives ##g##.
Is it correct? To me, it was necessary to proof also that ##g(g g^{-1}) = g##, because with this I could use the fact that the identity of a group is unique.
 

fresh_42

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To be exact, one first proves that left and right identity are the same: ##e_lg=ge_r=g \Longrightarrow e_l=e_r## and that there cannot be two identities, i.e. we have to show ##e=e'##. Then you can conclude that the right inverse is equal to the left inverse. (You wrote ##eg=ge## which you cannot know from the start, then you concluded ##gg^{-1}=e## which is also not known until the uniqueness of ##e## has been proven.)

Another way to define a group is by demanding that the equations ##ax=b## and ##xa=b## can uniquely be solved (IIRC).
 

WWGD

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A nice exercise in this area is to show that the identity of a subgroup must be the same as that of the supergroup. And that a subset S is a group iff for all a,b in S, ##ab^{-1}## is also in S.
 

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